If the probability that a fulurorescent light has a useful life of a least 800 hours is 0.9 find the probability that among 20 such lights at least 2 will not have a useful life of at least 800 hours. [ Given `(0.9) = 0.1348 ] `

Explanation

Let X = Number of fluorescent light which has useful life.
p = Probability that the light will not have useful life.
q = Probability that the light has useful life.
`:.q=0.9=(9)/(10)`
`:.p=1-q=1-(9)/(10)=(1)/(10)`
`n=20` (Given)
`:.X~B(n,p)`
`:.X~B(20,(1)/(10))`
The p.m.f. of X is given by
`P[X=x]=.^(n)C_(x)p^(x)q^(n-x.)`
i.e. ` P(x)=.^(20)C_(x)((1)/(10))^(x)*((9)/(10))^(20-x)`
`:.` P (at least 2 lights will not have a useful life )
`=P[X ge 2 ] = 1 - P [ X lt 2 ] `
`=1-[ P (X=0)+P(X=1)]`
`=1-[P(0)+P(1)]`
`=1-[.^(20)C_(0)((1)/(10))^(0)((9)/(10))^(20-0)+.^(20)C_(1)((1)/(10))^(1)((9)/(10))^(20-1)]`
`=1-[1xx1xx((9)/(10))^(20)+20xx(1)/(10)xx((9)/(10))^(19)]`
`=1-[(0.9)^(20)+2xx(0.9)^(19)]`
`=1-[(0.9+2)*(0.9)^(19)]`
`=1-[(2.9)(0.1348)]`
`=1-0.3910=0.6090`
Hence, the probability that at least 2 lights will not have a useful life of at least 800 hours is 0.6090.